Question

N.E.U. -Civil Engineering- CE241 Lecture Notes- Assist. Prof. Dr. Pinar Akpinar Ex-6:The dimensions of a steel element and the forces acting on it are, shown on the given figure. P1-6000kg DオP3-60000kgf P2-15000kgf AB-2cm, BC-1cm, CD-5cm a) What is the total change in volume of this steel element if the Modulus of Elasticity of steel is 2.1x 10 kgf/cm2 and Poisson's ratio is 0.25? b) What must be the magnitude of the compressive load acting in AB direction so that there will be no change in volume?

may you explain the way please

Answer 1

6_2 = 15000/1 times 5 = -3000 kgf/cm^2 6_1 = 6000/2 times 1 = 300 kgf/m^2 6_3 = 60000/2 times 5 = 6000 kgf/cm^2 volumetric strain E_v = E_T + E_2 + E_3 volumetric strain E_V = E_! + E_2 + E_3 E_1 = 6_1/E - mu 6_2/E - u 6_3/E 1/2.1 times 10^6[3000 + 0.25 times 3000 - 0.25 times 6000] = 3/2000 E_2 = 1/2.1 times 10^6[- 3000 - 0.25 times 3000 - 0.25 times 6000] = -2.5 times 10^- 3 E_3 = 1/2.1 times 10^6[6000 - 0.25 times 3000 + 0.25 times 300] = 1/350 therefore E_v = 2/2800 - 2.5 times 10^- 3 + 1/350 E +_v = 1/700 therefore change in volume = 1/700 times 2 times 1 times 5 = 0.0143 cm^3 E_v = 0 i.e: 1/2.1 times 10^6[3000 + 0.25 6_2 - 0.25 times 6000 - 6_2 - 0.25 times 3000 - 0.25 times times 6000 + 6000 - 0.25 times 3000 + 0.25 times 6_2] = -0.5 6_2 + 4500 = 0 6_2 = 9000 kgf/cm^2